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kNN and SVM

kNN and SVM. Michael L. Nelson CS 423/532 Old Dominion University. This work is licensed under a  Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. This course is based on Dr. McCown's class. Chapter 8. Some Evaluations Are Simple….

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kNN and SVM

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  1. kNN and SVM Michael L. Nelson CS 423/532 Old Dominion University This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License This course is based on Dr. McCown's class

  2. Chapter 8

  3. Some Evaluations Are Simple…

  4. Some Evaluations Are More Complex… • Chapter 8 price model for wines: price=f(rating,age) • wines have a peak age • far into the future for good wines (high rating) • nearly immediate for bad wines (low rating) • wines can gain 5X original value at peak age • wines go bad 5 years after peak age def wineprice(rating,age): peak_age=rating-50 # Calculate price based on rating price=rating/2 if age>peak_age: # Past its peak, goes bad in 5 years price=price*(5-(age-peak_age)) else: # Increases to 5x original value as it # approaches its peak price=price*(5*((age+1)/peak_age)) if price<0: price=0 return price

  5. To Anacreon in Heaven… def wineset1(): rows=[] for i in range(300): # Create a random age and rating rating=random()*50+50 age=random()*50 # Get reference price price=wineprice(rating,age) # Add some noise #price*=(random()*0.2+0.9) price*=(random()*0.4+0.8) # Add to the dataset rows.append({'input':(rating,age), 'result':price}) return rows >>> import numpredict >>> numpredict.wineprice(95.0,3.0) 21.111111111111114 >>> numpredict.wineprice(95.0,8.0) 47.5 >>> numpredict.wineprice(99.0,1.0) 10.102040816326529 >>> numpredict.wineprice(20.0,1.0) 0 >>> numpredict.wineprice(30.0,1.0) 0 >>> numpredict.wineprice(50.0,1.0) 112.5 >>> numpredict.wineprice(50.0,2.0) 100.0 >>> numpredict.wineprice(50.0,3.0) 87.5 >>> data=numpredict.wineset1() >>> data[0] {'input': (89.232627562980568, 23.392312984476838), 'result': 157.65615979190267} >>> data[1] {'input': (59.87004163297604, 2.6353185389295875), 'result': 50.624575737257267} >>> data[2] {'input': (95.750031143736848, 29.800709868119231), 'result': 184.99939310081996} >>> data[3] {'input': (63.816032861417639, 6.9857271772707783), 'result': 104.89398176429833} >>> data[4] {'input': (79.085632724279833, 36.304704141161352), 'result': 53.794171791411422} good wine, but not peak age = low price skunk water middling wine, but peak age = high price

  6. How Much is This Bottle Worth? • Find the k“nearest neighbors” to the item in question and average their prices. Your bottle is probably worth what the others are worth. • Questions: • how big should k be? • what dimensions should be used to judge “nearness”

  7. k=1, Too Small

  8. k=20, Too Big

  9. Define “Nearness” as f(rating,age) >>> data[0] {'input': (89.232627562980568, 23.392312984476838), 'result': 157.65615979190267} >>> data[1] {'input': (59.87004163297604, 2.6353185389295875), 'result': 50.624575737257267} >>> data[2] {'input': (95.750031143736848, 29.800709868119231), 'result': 184.99939310081996} >>> data[3] {'input': (63.816032861417639, 6.9857271772707783), 'result': 104.89398176429833} >>> data[4] {'input': (79.085632724279833, 36.304704141161352), 'result': 53.794171791411422} >>> numpredict.euclidean(data[0]['input'],data[1]['input']) 35.958507629062964 >>> numpredict.euclidean(data[0]['input'],data[2]['input']) 9.1402461702479503 >>> numpredict.euclidean(data[0]['input'],data[3]['input']) 30.251931245339232 >>> numpredict.euclidean(data[0]['input'],data[4]['input']) 16.422282108155486 >>> numpredict.euclidean(data[1]['input'],data[2]['input']) 45.003690219362205 >>> numpredict.euclidean(data[1]['input'],data[3]['input']) 5.8734063451707224 >>> numpredict.euclidean(data[1]['input'],data[4]['input']) 38.766821739987471

  10. kNN Estimator >>> numpredict.knnestimate(data,(95.0,3.0)) 21.635620163824875 >>> numpredict.wineprice(95.0,3.0) 21.111111111111114 >>> numpredict.knnestimate(data,(95.0,15.0)) 74.744108153418324 >>> numpredict.knnestimate(data,(95.0,25.0)) 145.13311902177989 >>> numpredict.knnestimate(data,(99.0,3.0)) 19.653661909493177 >>> numpredict.knnestimate(data,(99.0,15.0)) 84.143397370311604 >>> numpredict.knnestimate(data,(99.0,25.0)) 133.34279965424111 >>> numpredict.knnestimate(data,(99.0,3.0),k=1) 22.935771290035785 >>> numpredict.knnestimate(data,(99.0,3.0),k=10) 29.727161237156785 >>> numpredict.knnestimate(data,(99.0,15.0),k=1) 58.151852659938086 >>> numpredict.knnestimate(data,(99.0,15.0),k=10) 92.413908926458447 def knnestimate(data,vec1,k=5): # Get sorted distances dlist=getdistances(data,vec1) avg=0.0 # Take the average of the top k results for i in range(k): idx=dlist[i][1] avg+=data[idx]['result'] avg=avg/k return avg mo neighbors, mo problems

  11. Should All Neighbors Count Equally? • getdistances() sorts the neighbors by distances, but those distances could be: 1, 2, 5, 11, 12348, 23458, 456599 • We’ll notice a big change going from k=4 to k=5 • How can we weight the 7 neighbors above accordingly?

  12. Weight Functions Falls off too quickly Goes to Zero Falls Slowly, Doesn’t Hit Zero

  13. Weighted vs. Non-Weighted >>> numpredict.wineprice(95.0,3.0) 21.111111111111114 >>> numpredict.knnestimate(data,(95.0,3.0)) 21.635620163824875 >>> numpredict.weightedknn(data,(95.0,3.0)) 21.648741297049899 >>> numpredict.wineprice(95.0,15.0) 84.444444444444457 >>> numpredict.knnestimate(data,(95.0,15.0)) 74.744108153418324 >>> numpredict.weightedknn(data,(95.0,15.0)) 74.949258534489346 >>> numpredict.wineprice(95.0,25.0) 137.2222222222222 >>> numpredict.knnestimate(data,(95.0,25.0)) 145.13311902177989 >>> numpredict.weightedknn(data,(95.0,25.0)) 145.21679590393029 >>> numpredict.knnestimate(data,(95.0,25.0),k=10) 137.90620608492134 >>> numpredict.weightedknn(data,(95.0,25.0),k=10) 138.85154438288421

  14. Cross-Validation • Testing all the combinations would be tiresome… • We cross validate our data to see how our method is performing: • divide our 300 bottles into training data and test data (typically something like a (0.95,0.05) split) • train the system with our training data, then see if we can correctly predict the results in the test data (where we already know the answer) and record the errors • repeat n times with different training/test partitions

  15. Cross-Validating kNN and WkNN >>> numpredict.crossvalidate(numpredict.knnestimate,data) # k=5 357.75414919641719 >>> def knn3(d,v): return numpredict.knnestimate(d,v,k=3) ... >>> numpredict.crossvalidate(knn3,data) 374.27654623186737 >>> def knn1(d,v): return numpredict.knnestimate(d,v,k=1) ... >>> numpredict.crossvalidate(knn1,data) 486.38836851997144 >>> numpredict.crossvalidate(numpredict.weightedknn,data) # k=5 342.80320831062471 >>> def wknn3(d,v): return numpredict.weightedknn(d,v,k=3) ... >>> numpredict.crossvalidate(wknn3,data) 362.67816434458132 >>> def wknn1(d,v): return numpredict.weightedknn(d,v,k=1) ... >>> numpredict.crossvalidate(wknn1,data) 524.82845502785574 >>> def wknn5inverse(d,v): return numpredict.weightedknn(d,v,weightf=numpredict.inverseweight) ... >>> numpredict.crossvalidate(wknn5inverse,data) 342.68187472350417 In this case, we understand the price function and weights well enough to not need optimization…

  16. Heterogeneous Data • Suppose that in addition to rating & age, we collected: • bottle size (in ml) • 375, 750, 1500, 3000 • (book goes to 3000, code to 1500) • http://en.wikipedia.org/wiki/Wine_bottle#Sizes • the # of aisle where the wine was bought (aisle 2, aisle 9, etc.) def wineset2(): rows=[] for i in range(300): rating=random()*50+50 age=random()*50 aisle=float(randint(1,20)) bottlesize=[375.0,750.0,1500.0][randint(0,2)] price=wineprice(rating,age) price*=(bottlesize/750) price*=(random()*0.2+0.9) rows.append({'input': (rating,age,aisle,bottlesize), 'result':price}) return rows

  17. Vintage #2 >>> data=numpredict.wineset2() >>> data[0] {'input': (54.165108104770141, 34.539865790286861, 19.0, 1500.0), 'result': 0.0} >>> data[1] {'input': (85.368451290310119, 20.581943831329454, 7.0, 750.0), 'result': 138.67018277159647} >>> data[2] {'input': (70.883447179046527, 17.510910062083763, 8.0, 375.0), 'result': 83.519907955896613} >>> data[3] {'input': (63.236220974521459, 15.66074713248673, 9.0, 1500.0), 'result': 256.55497402767531} >>> data[4] {'input': (51.634428621301851, 6.5094854514893496, 6.0, 1500.0), 'result': 120.00849381080788} >>> numpredict.crossvalidate(knn3,data) 1197.0287329391431 >>> numpredict.crossvalidate(numpredict.weightedknn,data) 1001.3998202008664 >>> We have more data -- why are the errors bigger?

  18. Differing Data Scales distance=30 distance=180

  19. Rescaling The Data distance=18 Rescale ml by 0.1 Rescale aisle by 0.0

  20. Cross-Validating Our Scaled Data >>> sdata=numpredict.rescale(data,[10,10,0,0.5]) >>> numpredict.crossvalidate(knn3,sdata) 874.34929987724752 >>> numpredict.crossvalidate(numpredict.weightedknn,sdata) 1137.6927754808073 >>> sdata=numpredict.rescale(data,[15,10,0,0.1]) >>> numpredict.crossvalidate(knn3,sdata) 1110.7189445981378 >>> numpredict.crossvalidate(numpredict.weightedknn,sdata) 1313.7981751958403 >>> sdata=numpredict.rescale(data,[10,15,0,0.6]) >>> numpredict.crossvalidate(knn3,sdata) 948.16033679019574 >>> numpredict.crossvalidate(numpredict.weightedknn,sdata) 1206.6428136396851 my 2nd and 3rd guesses are worse than the initial guess -- but how can we tell if the initial guess is “good”?

  21. Optimizing The Scales >>> import optimization # using the chapter 5 version, not the chapter 8 version! >>> reload(numpredict) <module 'numpredict' from 'numpredict.pyc'> >>> costf=numpredict.createcostfunction(numpredict.knnestimate,data) >>> optimization.annealingoptimize(numpredict.weightdomain,costf,step=2) [4, 8.0, 2, 4.0] >>> optimization.annealingoptimize(numpredict.weightdomain,costf,step=2) [4, 8, 4, 4] >>> optimization.annealingoptimize(numpredict.weightdomain,costf,step=2) [6, 10, 2, 4.0] the book got [11,18,0,6] -- the last solution is close, but we are hoping to see 0 for aisle… code has: weightdomain=[(0,10)]*4 book has: weightdomain=[(0,20)]*4

  22. Optimizing - Genetic Algorithm >>> optimization.geneticoptimize(numpredict.weightdomain,costf,popsize=5) 1363.57544567 1509.85520291 1614.40150619 1336.71234577 1439.86478765 1255.61496037 1263.86499276 1447.64124381 [lots of lines deleted] 1138.43826351 1215.48698063 1201.70022455 1421.82902056 1387.99619684 1112.24992339 1135.47820954 [5, 6, 1, 9] the book got [20,18,0,12] on this one -- or did it?

  23. Optimization - Particle Swarm >>> import numpredict >>> data=numpredict.wineset2() >>> costf=numpredict.createcostfunction(numpredict.knnestimate,data) >>> import optimization >>> optimization.swarmoptimize(numpredict.weightdomain,costf,popsize=5,lrate=1,maxv=4,iters=20) >>> optimization.swarmoptimize(numpredict.weightdomain,costf,popsize=5,lrate=1,maxv=4,iters=20) [10.0, 4.0, 0.0, 10.0] 1703.49818863 [6.0, 4.0, 3.0, 10.0] 1635.317375 [6.0, 4.0, 3.0, 10.0] 1433.68139154 [6.0, 10, 2.0, 10] 1052.571099 [8.0, 8.0, 0.0, 10.0] 1286.04236301 [6.0, 6.0, 2.0, 10] 876.656865281 [6.0, 7.0, 2.0, 10] 1032.29545458 [8.0, 8.0, 0.0, 10.0] 1190.01320225 [6.0, 7.0, 2.0, 10] 1172.43008909 [4.0, 7.0, 3.0, 10] 1287.94875028 [4.0, 8.0, 3.0, 10] 1548.04584827 [8.0, 8.0, 0.0, 10.0] 1294.08912173 [8.0, 6.0, 2.0, 10] 1509.85587222 [8.0, 6.0, 2.0, 10] 1135.66091584 [10, 10.0, 0, 10.0] 1023.94077802 [8.0, 6.0, 2.0, 10] 1088.75216364 [8.0, 8.0, 0.0, 10.0] 1167.18869905 [8.0, 8.0, 0, 10] 1186.1047697 [8.0, 6.0, 0, 10] 1108.8635027 [10.0, 8.0, 0, 10] 1220.45183068 [10.0, 8.0, 0, 10] >>> included in chapter 8 code but not discussed in book! see: http://en.wikipedia.org/wiki/Particle_swarm_optimization cf. [20,18,0,12]

  24. Chapter 9

  25. Matchmaking Site male: female: age,smoker,wants children,interest1:interest2:…:interestN:addr, age,smoker,wants children,interest1:interest2:…:interestN:addr,match 39,yes,no,skiing:knitting:dancing,220 W 42nd St New York NY, 43,no,yes,soccer:reading:scrabble,824 3rd Ave New York NY,0 23,no,no,football:fashion,102 1st Ave New York NY, 30,no,no,snowboarding:knitting:computers:shopping:tv:travel,151 W 34th St New York NY,1 50,no,no,fashion:opera:tv:travel,686 Avenue of the Americas New York NY, 49,yes,yes,soccer:fashion:photography:computers:camping:movies:tv,824 3rd Ave New York NY,0 46,no,yes,skiing:reading:knitting:writing:shopping,154 7th Ave New York NY, 19,no,no,dancing:opera:travel,1560 Broadway New York NY,0 36,yes,yes,skiing:knitting:camping:writing:cooking,151 W 34th St New York NY, 29,no,yes,art:movies:cooking:scrabble,966 3rd Ave New York NY,1 27,no,no,snowboarding:knitting:fashion:camping:cooking,27 3rd Ave New York NY, 19,yes,yes,football:computers:writing,14 E 47th St New York NY,0 (linebreaks, spaces added for readability)

  26. Start With Only Ages… >>> import advancedclassify >>> matchmaker=advancedclassify.loadmatch('matchmaker.csv') >>> agesonly=advancedclassify.loadmatch('agesonly.csv',allnum=True) >>> matchmaker[0].data ['39', 'yes', 'no', 'skiing:knitting:dancing', '220 W 42nd St New York NY', '43', 'no', 'yes', 'soccer:reading:scrabble', '824 3rd Ave New York NY'] >>> matchmaker[0].match 0 >>> agesonly[0].data [24.0, 30.0] >>> agesonly[0].match 1 >>> agesonly[1].data [30.0, 40.0] >>> agesonly[1].match 1 >>> agesonly[2].data [22.0, 49.0] >>> agesonly[2].match 0 24,30,1 30,40,1 22,49,0 43,39,1 23,30,1 23,49,0 48,46,1 23,23,1 29,49,0 …

  27. M age vs. F age

  28. Not a Good Match For a Decision Tree(covered in Chapter 7)

  29. Boundaries are Vertical & Horizontal Only cf. L1 norm from ch 3; http://en.wikipedia.org/wiki/Taxicab_geometry

  30. Linear Classifier >>> avgs=advancedclassify.lineartrain(agesonly) avg. point for non-match avg. point for match Are (x,y) a match? Plot the data and compute which point is “closest”.

  31. Vector, Dot Product Review Instead of Euclidean distance, we’ll use vector dot products. A = (2,3) B = (-1,-2) AB = 2(-1) + 3(-2) AB = -8 also: AB = len(A)len(B)cos(AB) so: (X1-C)(M0-M1) is positive, so X1 is in class M0 (X2-C)(M0-M1) is negative, so X2 is in class M1 M0=match M1=no match

  32. Dot Product Classifier >>> avgs=advancedclassify.lineartrain(agesonly) >>> advancedclassify.dpclassify([50,50],avgs) 1 >>> advancedclassify.dpclassify([60,60],avgs) 1 >>> advancedclassify.dpclassify([20,60],avgs) 0 >>> advancedclassify.dpclassify([30,30],avgs) 1 >>> advancedclassify.dpclassify([30,25],avgs) 1 >>> advancedclassify.dpclassify([25,40],avgs) 0 >>> advancedclassify.dpclassify([48,20],avgs) 1 >>> advancedclassify.dpclassify([60,20],avgs) 1 these should not be matches!

  33. Categorical Features • Convert yes/no questions to: • yes = 1, no = -1, unknown/missing = 0 • Count interest overlaps. e.g., {fishing:hiking:hunting} and {activism:hiking:vegetarianism} will have an interest overlap of “1” • optimizations, such as creating a hierarchy of related interests, are desirable. • combining outdoor sports like hunting, fishing • if choosing from a bounded list of interests, measure the cosine between two resulting vectors • (0,1,1,1,0) (1,0,1,0,1) • if accepting free text from users, normalize the results • stemming, synonyms, normalize input lengths, etc. • Convert addresses to latitude, longitude, then convert lat,long pairs to mileage • mileage is approximate, but book has code with < 10% error which will be fine for determining proximity

  34. Yahoo Geocoding API >>> advancedclassify.milesdistance('cambridge, ma','new york,ny') 191.51092890345939 >>> advancedclassify.getlocation('532 Rhode Island Ave, Norfolk, VA') (36.887245, -76.286400999999998) >>> advancedclassify.milesdistance('norfolk, va','blacksburg, va') 220.21868849853567 >>> advancedclassify.milesdistance('532 rhode island ave., norfolk, va', '4700 elkhorn ave., norfolk, va') 1.1480170414890398 http://api.local.yahoo.com/MapsService/V1/geocode?appid=appid&location=532+Rhode+Island+Ave,Norfolk,VA 2013 update: of course this no longer works, but similar services are available

  35. Loaded & Scaled >>> numericalset=advancedclassify.loadnumerical() >>> numericalset[0].data [39.0, 1, -1, 43.0, -1, 1, 0, 6.729579883484428] >>> numericalset[0].match 0 >>> numericalset[1].data [23.0, -1, -1, 30.0, -1, -1, 0, 1.6738043955092503] >>> numericalset[1].match 1 >>> numericalset[2].data [50.0, -1, -1, 49.0, 1, 1, 2, 5.715074975686611] >>> numericalset[2].match 0 >>> scaledset,scalef=advancedclassify.scaledata(numericalset) >>> avgs=advancedclassify.lineartrain(scaledset) >>> scalef(numericalset[0].data) [0.65625, 1, 0, 0.78125, 0, 1, 0, 0.44014343540421147] >>> scaledset[0].data [0.65625, 1, 0, 0.78125, 0, 1, 0, 0.44014343540421147] >>> scaledset[0].match 0 >>> scaledset[1].data [0.15625, 0, 0, 0.375, 0, 0, 0, 0.10947399831631938] >>> scaledset[1].match 1 >>> scaledset[2].data [1.0, 0, 0, 0.96875, 1, 1, 0, 0.37379045600821365] >>> scaledset[2].match 0 >>> def loadnumerical(): oldrows=loadmatch('matchmaker.csv') newrows=[] for row in oldrows: d=row.data data=[float(d[0]),yesno(d[1]),yesno(d[2]), float(d[5]),yesno(d[6]),yesno(d[7]), matchcount(d[3],d[8]), milesdistance(d[4],d[9]), row.match] # [mAge,smoke,kids,fAge,smoke,kids,interest, # miles], match newrows.append(matchrow(data)) return newrows oldest couple, ages are 1.0 and 0.97; but age is the only thing going for them (I’m not sure why the interests were scaled to exactly 0; int vs. float error?)

  36. A Linear Classifier Won’t Help Idea: transform data… convert every (x,y) to (x2,y2)

  37. Now a Linear Classifier Will Help… That was an easy transformation, but what about a transformation that takes us to higher dimensions? e.g., (x,y)  (x2,xy,y2)

  38. The “Kernel Trick” • We can use linear classifiers on non-linear problems if we transform the original data into higher-dimensional space • http://en.wikipedia.org/wiki/Kernel_trick • Replace the dot product with the radial basis function • http://en.wikipedia.org/wiki/Radial_basis_function import math def rbf(v1,v2,gamma=10): dv=[v1[i]-v2[i] for i in range(len(v1))] l=veclength(dv) return math.e**(-gamma*l)

  39. Nonlinear Classifier >>> offset=advancedclassify.getoffset(agesonly) >>> offset -0.0076450020098023288 >>> advancedclassify.nlclassify([30,30],agesonly,offset) 1 >>> advancedclassify.nlclassify([30,25],agesonly,offset) 1 >>> advancedclassify.nlclassify([25,40],agesonly,offset) 0 >>> advancedclassify.nlclassify([48,20],agesonly,offset) 0 >>> ssoffset=advancedclassify.getoffset(scaledset) >>> ssoffset 0.012744361062728658 >>> numericalset[0].match 0 >>> advancedclassify.nlclassify(scalef(numericalset[0].data),scaledset,ssoffset) 0 >>> numericalset[1].match 1 >>> advancedclassify.nlclassify(scalef(numericalset[1].data),scaledset,ssoffset) 1 >>> numericalset[2].match 0 >>> advancedclassify.nlclassify(scalef(numericalset[2].data),scaledset,ssoffset) 0 >>> newrow=[28.0,-1,-1,26.0,-1,1,2,0.8] # Man doesn't want children, woman does >>> advancedclassify.nlclassify(scalef(newrow),scaledset,ssoffset) 0 >>> newrow=[28.0,-1,1,26.0,-1,1,2,0.8] # Both want children >>> advancedclassify.nlclassify(scalef(newrow),scaledset,ssoffset) 1 the dot product classfier (slide 32) predicted matches for these!

  40. Linear Misclassification

  41. Maximum-Margin Hyperplane H1 does not separate the classes at all. H2 separates the classes, but with a small margin. H3 separates the classes with the maximum margin. image from: http://en.wikipedia.org/wiki/Support_vector_machine

  42. Support Vector Machine Maximum-Margin Hyperplane Support Vectors

  43. Linear in Higher Dimensions original input dimensions higher dimensions via "kernel trick" image from: http://en.wikipedia.org/wiki/Support_vector_machine

  44. LIBSVM classes training data >>> from svm import * >>> prob = svm_problem([1,-1],[[1,0,1],[-1,0,-1]]) >>> param = svm_parameter(kernel_type = LINEAR, C = 10) >>> m = svm_model(prob, param) * optimization finished, #iter = 1 nu = 0.025000 obj = -0.250000, rho = 0.000000 nSV = 2, nBSV = 0 Total nSV = 2 >>> m.predict([1, 1, 1]) 1.0 >>> m.predict([1, 1, -1]) -1.0 >>> m.predict([0, 0, 0]) -1.0 >>> m.predict([1, 0, 0]) 1.0 check the errata for changes in the Python interface to libsvm: http://www.oreilly.com/catalog/errataunconfirmed.csp?isbn=9780596529321

  45. >>> answers,inputs=[r.match for r in scaledset],[r.data for r in scaledset] >>> param = svm_parameter(kernel_type = RBF) >>> prob = svm_problem(answers,inputs) >>> m=svm_model(prob,param) * optimization finished, #iter = 329 nu = 0.777729 obj = -290.207656, rho = -0.965033 nSV = 394, nBSV = 382 Total nSV = 394 >>> newrow=[28.0,-1,-1,26.0,-1,1,2,0.8]# Man doesn't want children, woman does >>> m.predict(scalef(newrow)) 0.0 >>> newrow=[28.0,-1,1,26.0,-1,1,2,0.8]# Both want children >>> m.predict(scalef(newrow)) 1.0 >>> newrow=[38.0,-1,1,24.0,1,1,1,2.8]# Both want children, but less in common >>> m.predict(scalef(newrow)) 1.0 >>> newrow=[38.0,-1,1,24.0,1,1,0,2.8]# Both want children, but even less in common >>> m.predict(scalef(newrow)) 1.0 >>> newrow=[38.0,-1,1,24.0,1,1,0,10.0]# Both want children, but far less in common, 10 miles >>> m.predict(scalef(newrow)) 1.0 >>> newrow=[48.0,-1,1,24.0,1,1,0,10.0]# Both want children, nothing in common, older male >>> m.predict(scalef(newrow)) 1.0 >>> newrow=[24.0,-1,1,48.0,1,1,0,10.0]# Both want children, nothing in common, older female >>> m.predict(scalef(newrow)) 1.0 >>> newrow=[24.0,-1,1,58.0,1,1,0,10.0]# Both want children, nothing in common, much older female >>> m.predict(scalef(newrow)) 1.0 >>> newrow=[24.0,-1,1,58.0,1,1,0,100.0]# Same as above, but greater distance >>> m.predict(scalef(newrow)) 0.0 LIBSVM on Matchmaker

  46. >>> guesses = cross_validation(prob, param, 4) * optimization finished, #iter = 206 nu = 0.796942 obj = -235.638042, rho = -0.957618 nSV = 306, nBSV = 296 Total nSV = 306 * optimization finished, #iter = 224 nu = 0.780128 obj = -237.590876, rho = -1.027825 nSV = 300, nBSV = 288 Total nSV = 300 * optimization finished, #iter = 239 nu = 0.794009 obj = -235.252234, rho = -0.941018 nSV = 307, nBSV = 289 Total nSV = 307 * optimization finished, #iter = 278 nu = 0.802139 obj = -234.473046, rho = -0.908467 nSV = 306, nBSV = 289 Total nSV = 306 >>> guesses [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, [much deletia] , 1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 0.0, 0.0] >>> sum([abs(answers[i]-guesses[i]) for i in range(len(guesses))]) 120.0 Cross-validation correct = 380/500 = 0.76 could we do better with different values for svm_parameter()?

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